The 1860 United States presidential election in Massachusetts took place on November 2, 1860, as part of the 1860 United States presidential election. Voters chose 13 electors of the Electoral College, who voted for president and vice president.
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| Turnout | 65.8%[1]  4.0 pp | ||||||||||||||||||||||||||||||||
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 County Results Lincoln 40-50% 50-60% 60-70% 70-80% 80-90%
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Massachusetts was won by Republican candidate Abraham Lincoln, who won the state by 42.57%.
With 62.80% of the popular vote, Massachusetts would prove to be Lincoln's third strongest state in the 1860 election in terms of popular vote percentage after neighboring Vermont and Minnesota.[2]
Results
| 1860 United States presidential election in Massachusetts[3] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Republican | Abraham Lincoln of Illinois | Hannibal Hamlin of Maine | 106,684 | 62.80% | 13 | 100.00% | ||
| Democratic | Stephen Arnold Douglas of Illinois | Herschel Vespasian Johnson of Georgia | 34,370 | 20.23% | 0 | 0.00% | ||
| Constitutional Union | John Bell of Tennessee | Edward Everett of Massachusetts | 22,331 | 13.15% | 0 | 0.00% | ||
| Southern Democratic | John Cabell Breckinridge of Kentucky | Joseph Lane of Oregon | 6,163 | 3.63% | 0 | 0.00% | ||
| N/A | Others | Others | 328 | 0.19% | 0 | 0.00% | ||
| Total | 169,876 | 100.00% | 13 | 100.00% | ||||